Finding intervals where the second derivative is positive
1 Answer
Lets compute $f''(x)$:
$$f'(x)=15x^4−30x^2.$$
\[f''(x)=60x^3-60x=60x(x^2-1)=60x(x+1)(x-1).\]
The roots of $f''(x)=60x(x+1)(x-1)$ are $x=0, x=1, x=-1$. Hence $f''>0$ on the following intervals
\[(1,\infty) \text{and} (-1,0).\]
We can also see that Hence $f''<0$ on the following intervals
\[(-\infty,-1) \text{and} (0,1).\]

574
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- 1 Answer
- 348 views
- Pro Bono
Related Questions
- Recursive square root sequence
- Integral of trig functions
- Help with Norms
- Is $\sum_{i=1}^{\infty}\arctan (\frac{n+1}{n^2+5})$ convergent or divergent?
- Solve the attached problem
- A rectangular garden plot is to be fenced off along the property line.
- Evaluate $\int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx$
- Differentiate $f(x)=\int_{\tan x}^{0} \frac{\cos t}{1+e^t}dt$